AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be an acute triangle with 
, midpoint 
of side 
, altitude 
(
), and orthocenter 
. A circle passes through points 
and 
, is tangent to line 
, and intersects the circumcircle of triangle at a second point 
. The circumcircle of triangle 
intersects line 
at a second point 
. Prove that the lines 
and 
are perpendicular.
satisfy,
Where {
} is a permutation of . Prove that 
and 
Prove that
Let and 
Prove that
has a unique solution in integers with 
and 
.
as 
fir 
. Prove that 
for all 
.
be triangle with circumcenter 
and orthocenter 
. 
intersect again at 
respectively. Lines through parallel to 
intersects 
at 
respectively. Point 
such that 
is a parallelogram. Prove that lines 
and 
are concurrent at a point on 
.
is a cyclic quadrilateral inscribed in a circle of radius one unit. If 
, prove that is a square.
are positive real numbers.Prove that
be a function taking in positive integers and outputting nonnegative integers, defined as follows:
is the number of positive integers 
with 
such that the equation 
has an integer solution 
.
such that
and 
(Adit Aggarwal, India.)
be positive reals such that 
. Prove that
be an isosceles trapezoid with 
. The inscribed circle 
of triangle 
meets 
at 
. Let 
be a point on the (internal) angle bisector of 
such that 
. Let the circumscribed circle of triangle 
meet line at 
and 
. Prove that the triangle 
is isosceles.
such that, for all positive integers 
and ,
to deduce all of them are not divisible by 
denote the k-th symmetric sum and 
that 
and by newton sums we have 
and we consider the polynomial
which in 
is 
and the roots are 
so we get 
but 
and we are done
we get 
so as 
we have 
as desired.
, consider the polynomial 
, clearly 
and ![$P \in \mathbb Z[x]$](http://latex.artofproblemsolving.com/a/0/5/a05997bd7ba17001c62bc17ae35c76dd78e24789.png)
. And therefore we can let 
where all 
's are integers, so by taking newton sums modulo we have 
which gives that 
therefore by putting 
in this we get that 
for any 
which proves that all 's are congrent modulo and therefore we are done 
.
. Suppose 
of the 's are nonzero and WLOG they are 
. Let 
be a generator of 
. Let 
where 
is an integer for 
. Then the condition becomes 
for all 
. It follows that the elements of are roots of ![$P(x):=x^{\alpha_1}+\cdots+x^{\alpha_r}\in\mathbb{F}_p[x]$](http://latex.artofproblemsolving.com/8/b/0/8b0a5b5937b0aff17391fefd8df1603fb6c521a8.png)
. Thus 
is divisible by![\[
\prod_{a\in\mathbb{F}_p^\times}(x-a)=x^{p-1}-1.
\]](http://latex.artofproblemsolving.com/c/9/d/c9d1ccf7462e298113392f4798239e5f502ae2f4.png)
Since 
if 
is nonzero, 
and thus either 
in which case we are done, or 
and all the 
's are equal, as desired. 
be reals such that 
Prove that
be a prime number and 
be integers for which 
is divisible by 
.
is divisible by 
)
denote the 
-
,
.
(interpreted in ![$\mathbb F_p[t]$](http://latex.artofproblemsolving.com/d/1/f/d1fa992d9fe693e162a54b2121d646af3dd9f0e2.png)
)
,
.
.
But polynomials in 
modulo